MHT CET 2020 Physics Solved Paper

For the organ pipe P1 closed at one end the frequency is
v1=(2n+1)‌

v

4l1

where, variables have their usual meanings.
The first overtone frequency will obtain for n=1 ,
v1=(2(1)+1)‌

v

4l1

v1=‌

3v

4l1

...
For the organ pipe P2 open at both ends, the frequency is
v2=‌

nv

2l2

where, variables have their usual meanings.
The first overtone frequency will obtain for n=2,
hence, the third overtone frequency will obtain for n=4,
v2=‌

4v

2l2

These frequencies are in a resonance with a given tuning fork. Hence,
v1=v2
‌

3v

4l1

=‌

4v

2l2

‌

l1

l2

=‌

6

16

‌

l1

l2

=‌

3

8